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Chapter 3
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* Likelihood that a product will form when a substrate molecule enters
the active site
* Sometimes, with low enzyme affinity, a substrate will enter an
active site but instead of making a product, substrate will leave the
active site before a reaction takes place and a product forms
*
* The rate at which an enzyme can
convert a substrate molecule into
product per second
* Relies heavily on amount of
collisions between enzyme and
substrate molecule
* Maximum number of chemical
conversions of substrate
molecules per second that a
single catalytic site will execute
for a given enzyme concentration
* “Kcat” in formulas
*
* Vmax  theoretical maximum rate
of an enzyme ; tells us how well an
enzyme performs
* At Vmax  ALL enzyme molecules
are bound to substrate molecule
* Enzyme is SATURATED with substrate
* How to Measure Vmax:
* Reaction rate is measured at
different substrate concentrations
while keeping enzyme concentration
constant
* As substrate concentration
INCREASES reaction rate rises until
it reaches maximum rate (Vmax)
* Vmax is where your graph levels
off…
* Find the y-value for the plateau…this
is Vmax
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* Plotted against
substrate
concentrations
* Produces ASYMPTONIC curve
* In practice:
* Curve never truly flattens out
* In theory:
* Curve flattens out when there is an
infinite substrate concentration
* Impossible to read a value of Vmax
off a graph…so we must use another
graph….
*
* Create a double reciprocal plot (aka
Lineweaver-Burke plot)
* Plot inverse of substrate concentration on
x-axis (1/[S])
* Plot inverse of velocity on y-axis
(1/velocity)
* Benefits:
* 1/infinity substrate concentration = ZERO
* Zero can be plotted on your graph
* Vmax can be accurately found off of graph
* Resulting line is a straight line (not curve)
* Accurately determine Vmax
* Accurately determine Km (Michaelis-Menton
Constant)
* Steps for Finding Vmax from a Double-Reciprocal
Plot
* Find 1/Vmax
* Point where the line crosses/intersects the y-axis
* Point of intersection on y-axis is where 1/[S] is equal to zero
(when [S] is infinity)
* Once 1/Vmax is determined, Vmax can be calculated
* Km (Michaelis Menten Constant) can also be determined
from the double reciprocal plot
*
* Km (abbreviation)
* Measure of the affinity of an enzyme
• Higher affinity= lower
Michaelis-Menton constant =
for its substrate
reaction will occur much
more quickly (towards its
* The substrate concentration at which
an enzyme works at HALF its maximum
Vmax)
rate (1/2 Vmax)
• Lower affinity = higher
* At this point, HALF the active sites of
Micahelis-Menton constant =
the enzymes are occupied by the
reaction will occur much
substrate
more slowly
* Higher affinity of the enzyme for the
substrate  lower concentration is
needed for Km
• Low Km = High affinity of
enzyme for substrate =
more product
• High Km = Low affinity of
enzyme for substrate =
less product
*
* Use double reciprocal plot
* -1/Km is the point where the line for the graph
intersects x-axis
* Use the value of -1/Km to calculate Km
* Value of Km can vary depending on many factors
* Identity of substrate
* Temperature
* pH
* Presence of particular ions
* Overall ion concentration
* Presence of poisons
* Presence of pollutants
* Presence of inhibitors
* Vmax
*
* Info about maximum rate of
reaction that is possible
* Km
* Measures affinity of the enzyme
for the substrate
* Higher affinity = more likely
product will be formed when
substrate enters active site
* High affinity produces LOW Km
value
* Low affinity produces HIGH Km
value…
* What would give you a HIGH Km
value (low affinity for substrate to
go to active site)?
* INHIBITORS!!!!
*
* The smaller the 1/Km value is  larger the Km value is
* The larger the 1/Km value is  smaller Km value is
* The smaller the 1/Vmax value is  larger the Vmax value is
* The larger the 1/Vmax value is  smaller Vmax value is
Large Vmax means you need to add lots a substrate to make reaction
happen faster
Small Vmax  means only a little substrate is need to make reaction happen
fast
Large Km value LOW enzyme affinity (some reason, the substrates do NOT
want to bind to active site)
Low Km Value HIGH enzyme affinity (substrates immediately bind to
active sit of enzymes…only a small substrate concentration is needed to
saturate your enzymes’ active sites)
* Steps:
* Create table recording “v” (velocity) vs.
substrate concentration [S]
* Plot rate of reaction “v” (velocity) vs. substrate
concentration [S]
* V  y-axis
* [S] x-axis
* Can’t tell Vmax from graph….
* Revisit table and now fill in additional columns,
adding 1/v and 1/[S] by doing simple calculations
* Now, make a new graph
* Plot 1/v on y-axis and 1/[S] on x-Axis
* Plot values from table on this new graph
* Draw straight line through points….ALL the way
through y-axis and onto negative side of x-axis
* Your y-intercept is the 1/Vmax when the 1/[S] is
0 (which, do the math, would equal INFINITY!!!!
Endless substrate, yay!)
* Your x-intercept is the 1/km
* Plug in these values into the above equations,
solve for Km and Vmax, now you have
determined Vmax for your original graph!!!
[S]/
1/[S]/ v/
1/v /
arbitrary
unit
arbitrary
unit
arbitrary
unit
arbitrary
unit
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